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PSYCHO15rus [73]
3 years ago
8

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

Physics
2 answers:
salantis [7]3 years ago
5 0

Answer:

130m

Explanation:

You just have to multiply velocity by the time traveled:

100m/s * 1.3s = 130m!

juin [17]3 years ago
3 0

<u>Answer:</u>

The bullet would travel 130 m in 1.3 sec

<u>Solution: </u>

The bullet leaves a gun with a horizontal velocity of 100 m/s

We know that the velocity means the distance travelled per sec

The formula for the distance with respect to speed and time is

Distance = velocity \times time

Let us assume the bullet travelled distance S and the time is 1.3 sec

So, S = (100 \times1.3) m = 130 m

Hence, the bullet travel 130 m in 1.3 sec

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Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap
eduard

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

4 0
3 years ago
Read 2 more answers
Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?
Katen [24]

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

6 0
1 year ago
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially
ivanzaharov [21]

Answer:

a) s,f,r  b) r c) f

Explanation:

To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged

For the sphere to be in equilibrium the sum of forces is zero

    B - W = 0

    B = W = mg

Now let's use the concept of density for the body and water

Solid sphere

   ρ = m / V

  V = 4/3 π r³

   m = ρ₀ (4/3 π r³)

   W = ρ₀ (4/3 π r³) g

Water  (a)

   ρ = mₐ / Vₐ

   mₐ = ρ Vₐ

   B = ρ Vₐ g

Let's replace and simplify

   ρ Vₐ g = ρ₀ (4/3 π r³) g

    ρ Vf = ρ₀ (4/3 π r³)            (1)

For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

   r  <ro        V <V₀   ⇒      ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

Assume the three possibilities

- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)

- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)

- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

b) sphere the same radius, but the density increases.

In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

      r = r₀     V = V₀     m <m₀        ρ₁ <ρ₀

The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

8 0
2 years ago
How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?
OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹  electrons

= 1.90×10²⁰ Electrons

8 0
3 years ago
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
2 years ago
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